Answer:
i) After 20 s the concentration of C is 0.024 M
ii) after 15 min the concentration of C is 0.027 M
Explanation:
Let´s assume that the reaction is first order for each reactive, making a second order global reaction. In that case:
v = k[A][B] = 0.34 M⁻¹ s⁻¹ * 0.027 M * 0.130 M = 1.2 x 10⁻³ M /s
First, let´s see how much time it takes for the reactants to disappear:
The rate of disappearance of A will be:
-Δ[A] / Δt = v
where:
Δ[A] = final - initial concentration of A
Δ[t] = elapsed time
Then:
Δ[t] = -Δ[A] / v
Δ[t] = - (-0.027 M) / 1.2 x 10⁻³ M /s = 22.5 s
The rate of disappearance of B will be:
-1/2Δ[B] / Δt = v
Δt = -1/2 * (-0.130 M) / 1.2 x 10⁻³ M /s
Δt = 54.2 s
Then, the reactant A will completely disappear in 22.5 s and it will limit the reaction.
The rate of production of C will be:
Δ[C] / Δt = v
where
Δ[C] = final concentration of C - initial concentration of C
Δt = final time - initial time
v = rate of the reaction
Then:
Δ[C] = v * Δt
Since the initial concentration of C is 0 and the initial time is 0, then:
[C] = v * t
i) t = 20 s
[C] = 1.2 x 10⁻³ M /s * 20 s = <u>0.024 M</u>
ii) t = 15 min = 15 min * (60 s/ 1 min) = 900 s
The reaction does not occur beyond the 22.5 s which is the time in which A disappears. Then, the concentration of C at 15 min will be the same as the concentration at 22.5 s:
[C] = 1.2 x 10⁻³ M /s * 22.5 s = <u>0.027 M</u>
